CBSE Class 9 Answered
Expert Answer
Table A:
|
Height from which an object is dropped ‘h’ (m) |
Velocity of an object falling from the height ‘v2’ (m/s2) |
Potential energy mgh (J) |
Kinetic energy ½mv2 (J) |
TE = PE + KE |
|
10 |
0 |
20 × 10 ×10 = 2000 |
½ × 20 × 0 = 0 |
2000 |
|
2 |
80 |
20 × 10 ×2 = 400 |
½ × 20 × 80 = 800 |
1200 |
Table B:
|
Height from which an object is dropped ‘h’ (m) |
Velocity of an object falling from the height ‘v2’ (m/s2) |
Potential energy mgh (J) |
Kinetic energy ½mv2 (J) |
TE = PE + KE |
|
10 |
0 |
20 × 10 × 10 = 2000 |
½ × 20 × 0 = 0 |
2000 |
|
2 |
40 |
20 × 10 × 2 = 400 |
½ × 20 × 40 = 400 |
800 |
According to the law of conservation of energy, the sum of the potential energy and kinetic energy of the object should remain the same at every point during its fall.
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